Problem: $A=\left[\begin{array}{rr}6 & 13 & -7 & 9 \\3 & 3 & 1 & 4 \\6 & 2 &-8 & 19 \\2 &6 &8 & 9\end{array}\right]$ $A_{4,1}=$
Explanation: Background An $m\times n$ matrix has $m$ rows and $n$ columns. $A=\left[\begin{array}{rr}A_{1,1} & \cdots & A_{1,n} \\\\\vdots \ & \ddots & \vdots \\\\A_{m,1} &\cdots &A_{m,n}\end{array}\right]$ Therefore, the entry $A_{{c},{d}}$ is located on row ${c}$ and column ${d}$. Finding $A_{4,1}$ $A_{{4},{1}}$ is located on row ${4}$ of $A$ : $\left[\begin{array}{rr}6 & 13 & -7 & 9 \\3 & 3 & 1 & 4 \\6 & 2 &-8 & 19 \\ {2} & {6} & {8} & {9}\end{array}\right]$ $A_{{4},{1}}$ is also located on column ${1}$ of $A$. $\left[\begin{array}{rr}6 & 13 & -7 & 9 \\3 & 3 & 1 & 4 \\6 & 2 &-8 & 19 \\ {\text{2}} & {6} & {8} & {9}\end{array}\right]$ Therefore, $A_{{4},{1}}={2}$. Summary $A_{4,1}=2$